Question

33) A media analyst is trying to determine the proportion of people living in St. Louis...

33)

A media analyst is trying to determine the proportion of people living in St. Louis that watch a particular television show. The analyst collects a random sample of 268 people from St. Louis. Of the 268 people in the sample, 53 people said that they did watch the show.

a) Find a 95% confidence interval for the true population proportion of people in St. Louis who watch the show.

b) Provide the right endpoint of the interval as your answer.

Round your answer to 4 decimal places.

Please explain each step in detail (even if you are using a ti-84 calculator, please list steps!) Thank you!

Homework Answers

Answer #1

Let p denotes the true population proportion of people in St. Louis who watch the show.

a)

b) The right endpoint of the interval = 0.2454

To calculate z0.025 , I used R software and command : qnorm(1 - 0.025, 0, 1) .

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